Download Lots of proofs (answers to many worksheets)

Exercises:
1. Write a general way to express each of the following:
A) a rational number: a/b where a and b are integers and b ≠ 0
B) an even integer: 2k where k  Integers
C) an odd integer: 2k + 1 where k  Integers
D) a complex number: a + bi where a, b  Reals and i =
1
II. A) Prove even integers are closed with respect to addition.
Let a and b be even integers. Then a = 2k and b = 2m, where k, m  I. The sum of
a and b = a + b = 2k + 2m = 2(k + m). Since I are closed with respect to addition,
k + m  I and that means a + b = 2(k + m) is even. Hence, even integers are closed
with respect to addition.
B) Prove odd integers are closed with respect to multiplication.
Let a and b be odd integers. Then a = 2k + 1 and b = 2m + 1 , where k, m  I.
The product of a and b = ab = (2k + 1)(2m + 1) = 4km + 2k + 2m + 1 =
2(2km + k + m) + 1. Since I are closed with respect to addition and multiplication,
2km + k + m  I and that means ab = 2(2km + k + m) + 1 is odd. Hence, odd integers
are closed with respect to multiplication.
C) Prove rational numbers are closed with respect to addition.
Let a and b be rational numbers. Then a = m/n and b = p/q, where m, n, p and q  I
And nq ≠ 0. The sum of a and b is a + b = m/n + p/q = (mq + np)/(nq). Since I are
closed wrt addition and multiplication, mq + np  I and nq  I. Furthermore nq ≠ 0.
Therefore a + b = (mq + np)/(nq) is rational. Therefore rational numbers are closed
with respect to addition.
D) Prove complex numbers are closed wrt multiplication.
Let a and b be complex numbers. Then a = r + si and b = t + ui, where r, s, t, u  R
and i = 1 . The product of a and b is ab = (r + si)(t + ui) = rt + rui + sti + sui2 =
= rt + rui + sti – su = (rt – su) + (ru + st)i . Since R are closed wrt multiplication and
addition, rt – su  R and ru + st R. That means ab = (rt – su) + (ru + st)i  C.
Therefore complex numbers are closed with respect to multiplication.
E) Prove the sum of two odd integers is even.
Let a and b be odd integers. Then a = 2k + 1 and b = 2m + 1 , where k, m  I.
The sum of a and b = ab = (2k + 1) + (2m + 1) = 2k + 2m + 2 = 2(k + m + 1). Since
I are closed with respect to addition and multiplication, k + m + 1 I and that
means ab = 2(k + m + 1) is even. Hence, the sum of two odd integers is even.
Fun with Groups!
1. Closure holds because every element in the body of the table is an element of
{a, b, c, d}. It would take 64 cases to prove associativity, so I’ll check 3 cases and base
my conclusion on that. Does:
(a * b) * c = a * (b * c)?
(c * d) * b = c * (d * b)
(d * c) * a = d * (c * a)
b*c =a*d
b*b=c*a
b*a=d*c
d =d 
c=c 
b=b
Based on this evidence, I will assume that the associative property holds. Since for each
element k of this set (Note that k is a variable that represents any element {a, b, c, d}),
a * k = k * a = k, a is the identity element for *. Since a * a = a, b * d = d * b = a and c * c
= a, each element of {a, b, c, d} has an inverse. Therefore, I conclude that ({a, b, c, d}, *)
forms a group.
2. Closure fails because d * b = e and e {a, b, c, d}.
3. It would take 64 cases to prove associativity, so I’ll check 3 cases and base my
conclusion on that. Does:
(a * b) * c = a * (b * c)?
(c * d) * b = c * (d * b)
(d * b) * a = d * (b * a)
d*c =a*b
d*b=c*a
a*a=d*b
d =d 
a=a 
c≠a 
Since the last example wasn’t true, the associativity property fails.
4. Since there is no element g, so that g * b = b * g = b, there is no identity element for
*.
5. Although d might be considered an identity (although for a and b it is not unique, so
technically it isn’t) b has no inverse, for there is no element k such that b * k = k * b = d.
This means that the inverse property fails (as well as the identity).
6. Does (x * y) * z = x * (y * z)
(xy + 2) * z = x * (yz + 2) Since these are not equivalent expressions,
(xy + 2)z + 2 = x(yz + 2) + 2 the associative property does not hold.
xyz + 2z + 2 = xyz + 2x + 2
7. If a and b  R, a * b = 4ab  R because 4  R and R are closed with respect to
multiplication. Similarly, b * a = 4 ba R. This means that closure holds. Does
(x * y) * z = x * (y * z)
(4xy) * z = x * (4yz) Since these are equivalent expressions,
4(4xy)z = 4x(4yz) the associative property holds.
16xyz = 16xyz
Is there an identity element g for which a * g = g * a = a for all a  R?
a * g = g * a = a  4ag = 4ga = a  4ag = a  4g = 1  g = ¼ . Since ¼  R, ¼ is the
identity element for *. If each element has an inverse, then for all a  R, there exists an
a-1  R, so that a * a-1 = a-1 * a = g  4aa-1 = 4a-1a = ¼  4aa-1 = ¼  a-1 = 1/(16a). Since,
if a  R, 1/(16a)  R, every real number has an inverse under *. In conclusion, (R, *) is
closed, associative, has an identity element of ¼ and each real has an inverse under *, so
(R, *) forms a group.
8. Is there an identity element g for which a * g = g * a = a for all a  Q?
a * g = g * a = a  a - 4g = 4g - a = a 
a – 4g = a  -4g = 0  g = 0 .
(Since these expressions aren’t equal)
4g – a = a  4g = 2a  g = a/2.
Since 0 ≠ a/2 and since a/2 isn’t a unique number, * can’t claim an identity element.
(Associativity also failed.)
9. If a and b  C, a * b = a + b + 3  C because 3  C and C are closed with respect to
addition. Similarly, b * a = b + a + 3  C. This means that closure holds. Does
(x * y) * z = x * (y * z)
(x + y + 3) * z = x * (y + z + 3) Since these are equivalent expressions,
(x + y + 3) + z + 3 = x + (y + z + 3) + 3 the associative property holds.
x+y+z+6=x+y+z+6
Is there an identity element g for which a * g = g * a = a for all a  C?
a * g = g * a = a  a + g + 3 = g + a + 3 = a  a + g + 3 = a  g = -3 . Since -3  C, -3 is
the identity element for *. If each element has an inverse, then for all a  C, there
exists an a-1  C, so that a * a-1 = a-1 * a = g  aa-1 + 3 = a-1a + 3 = -3 
aa-1 + 3 = -3  a-1 = -6 - a. Since, if a  C, -6 - a  C, every complex number has an
inverse under *. In conclusion, (C, *) is closed, associative, has an identity element of -3
and each real has an inverse under *, so (C, *) forms a group.
10. This proof is similar to #9 except g = 4 and a-1 = 8 – a. This does form a group.
11. Is there an identity element g for which a * g = g * a = a for all a  C?
a * g = g * a = a  6a/g = 6g/a = a 
6a/g = a  g = 6 .
(Since these expressions aren’t equal)
6g/a = a  g = a2/6 .
Since 6 ≠ a2/6 and since a2/6 isn’t a unique number, * can’t claim an identity element.
(Associativity also failed.)
More Proofs and Bases:
II. 1. Multiples of 5 are closed with respect to addition.
Let a and b be multiples of 5. Then a = 5k and b = 5m, where k, m  I. a + b =
5k + 5m = 5(k + m). Since I are closed wrt addition, k + m  I and so a + b =
5(k + m) is a multiple of 5. Hence, multiples of 5 are closed with respect to
addition.
2. If x is an integer that is not a multiple of 3, then x2 has a remainder of 1 when
divided by 3. (Please note that the phrase “is an integer that” has been added to
limit the number of cases.)
If x is an integer that is not a multiple of 3, then either x = 3k + 1 or x = 3k + 2
where k  I. (i.e. There is a remainder of 1 or 2 when x is divided by 3.)
Consider case #1: If x = 3k + 1, then x2 = (3k + 1)2 = 9k2 + 6k + 1 = 3(3k2 + 2k) + 1.
Since 3, 2, k  I and since I are closed wrt addition and multiplication, 3k2 + 2k  I
meaning x2 has a remainder of 1 when divided by 3.
Consider case #2: If x = 3k + 2, then x2 = (3k + 2)2 = 9k2 + 12k + 4 =
3(3k2 + 4k + 1) + 1. Since 3, 4, 1, k  I and since I are closed wrt addition and
multiplication, 3k2 + 4k + 1  I meaning x2 has a remainder of 1 when divided by 3.
In conclusion, considering all possible cases, if x is an integer that is not a multiple
of 3, then x2 has a remainder of 1 when divided by 3.
3. Integer multiples of any integer k are closed with respect to multiplication.
Let a and b be multiples of k. Then a = km and b = kn, where k, m, n  I. ab =
kmkn = k(kmn). Since I are closed wrt addition, kmn  I and so ab = k(kmn) is a
multiple of k. Hence, multiples of any integer k are closed with respect to
multiplication.
4. If d10 = abk where {d, a, b, k}  Digits, then k < 10. If k  Digits, then k < 10. Duh!
5. If x3 is not an odd integer, then x is not an odd integer. (We will assume the
negation of the conclusion and prove the negation of the hypothesis, thus proving
the contrapositive of the statement.) If x is an odd integer, then x = 2k + 1 where
k  I. x3 = (2k + 1)3 = 8k3 + 12k2 + 6k + 1 = 2(4k3 + 6k2 + 3k) + 1. Since
{4, 6, 3, k}  I and I are closed wrt addition and multiplication, 4k3 + 6k2 + 3k  I
And x3 = 2(4k3 + 6k2 + 3k) + 1 is odd. So if x is an odd integer, then x3 is an odd
integer. Furthermore, the contrapositive which is “if x3 is not an odd integer, then
x is not an odd integer” is also true.
6. If |m – n| < p and if n, p > 0, then n – p < m < n + p (Note the typo!)
Let m, n, p be real numbers so that n, p < 0. If |m – n| < p, then 2 cases are
possible:
Consider case #1: m – n > 0 and m – n < p. This means that m > n and m < n + p.
Since n, p > 0, n + p > n, so the intersection of these sets is n < m < p + n or
m  [n, p + n).
Consider case #2: m – n < 0 and –(m – n) < p. This means that m < n and m – n > -p
which also means that m > n - p. Since n, p > 0, n – p < n, so the intersection of
these sets is n – p < m < n or m  (n – p, n).
The union of case 1 and case 2 is that m  (n – p, n + p) or n – p < m < n + p